Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 10010;
int a[MAXN];
int b[MAXN];
int n;
vector<pair<int, int> > qa, qb;
void Xor(int *a, int i, int j, vector<pair<int, int> > &q) {
  q.push_back(make_pair(i, j));
  a[i] ^= a[j];
}
void guass(int *a, vector<pair<int, int> > &q) {
  int p = 0;
  for (int i = 29; i >= 0; i--) {
    for (int j = p; j < n; j++) {
      if (a[j] & (1 << i)) {
        if (j != p) {
          Xor(a, p, j, q);
          Xor(a, j, p, q);
          Xor(a, p, j, q);
        }
        for (int k = 0; k < n; k++) {
          if (k != p && a[k] & (1 << i)) {
            Xor(a, k, p, q);
          }
        }
        ++p;
        break;
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  guass(a, qa);
  guass(b, qb);
  for (int i = 0; i < 30 && i < n; i++) {
    if (a[i] == 0) break;
    int k = (1 << 29);
    while (!(k & a[i])) k >>= 1;
    for (int j = 0; j < 30 && j < n; j++) {
      if ((a[j] ^ b[j]) & k) {
        Xor(a, j, i, qa);
      }
    }
  }
  int ok = 1;
  for (int i = 0; i < n; i++)
    if (a[i] != b[i]) {
      ok = 0;
      break;
    }
  if (!ok)
    printf("-1\n");
  else {
    printf("%d\n", qa.size() + qb.size());
    for (int i = 0; i < qa.size(); i++)
      printf("%d %d\n", qa[i].first + 1, qa[i].second + 1);
    reverse(qb.begin(), qb.end());
    for (int i = 0; i < qb.size(); i++)
      printf("%d %d\n", qb[i].first + 1, qb[i].second + 1);
  }
}